CBSE NCERT Solutions for Class 12 Maths Chapter 05...Class-XII-Maths Continuity and...
Transcript of CBSE NCERT Solutions for Class 12 Maths Chapter 05...Class-XII-Maths Continuity and...
-
Class-XII-Maths Continuity and Differentiabil ity
1 Practice more on Continuity and Differentiability www.embibe.com
CBSE NCERT Solutions for Class 12 Maths Chapter 05
Back of Chapter Questions
Exercise 𝟓. 𝟏
1. Prove that the function 𝑓(𝑥) = 5𝑥 − 3 is continuous at 𝑥 = 0, at 𝑥 = −3 and at 𝑥 = 5.
Solution:
Given function is 𝑓(𝑥) = 5𝑥 − 3
At 𝑥 = 0, 𝑓(0) = 5(0) − 3 = −3
LHL = lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
(5𝑥 − 3) = −3
RHL = lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
(5𝑥 − 3) = −3
Here, at 𝑥 = 0, LHL = RHL = 𝑓(0) = −3
Hence, the function 𝑓 is continuous at 𝑥 = 0.
Now at 𝑥 = −3, 𝑓(−3) = 5(−3) − 3 = −18
LHL = lim𝑥→−3−
𝑓(𝑥) = lim𝑥→−3−
(5𝑥 − 3) = −18
RHL = lim𝑥→−3+
𝑓(𝑥) = lim𝑥→−3+
(5𝑥 − 3) = −18
Here, at 𝑥 = −3, LHL = RHL = 𝑓(−3) = −18
Hence, the function 𝑓 is continuous at 𝑥 = −3.
At 𝑥 = 5, 𝑓(5) = 5(5) − 3 = 22
𝐿𝐻𝐿 = lim𝑥→5−
𝑓(𝑥) = lim𝑥→5−
(5𝑥 − 3) = 22
RHL = lim𝑥→5+
𝑓(𝑥) = lim𝑥→5+
(5𝑥 − 3) = 22
Here, at 𝑥 = 5, LHL = RHL = 𝑓(5) = 22
Hence, the function 𝑓 is continuous at 𝑥 = 5.
2. Examine the continuity of the function 𝑓(𝑥) = 2𝑥2 − 1 at 𝑥 = 3.
-
Class-XII-Maths Continuity and Differentiabil ity
2 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given function is 𝑓(𝑥) = 2𝑥2 − 1.
At 𝑥 = 3, 𝑓(3) = 2(3)2 − 1 = 17
LHL = lim𝑥→3−
𝑓(𝑥) = lim𝑥→3−
(2𝑥2 − 1) = 17
RHL = lim𝑥→3+
𝑓(𝑥) = lim𝑥→3+
(2𝑥2 − 1) = 17
Here, at 𝑥 = 3, LHL = RHL = 𝑓(3) = 17
Therefore, the function 𝑓 is continuous at 𝑥 = 3.
3. Examine the following functions for continuity:
(a) 𝑓(𝑥) = 𝑥 − 5
(b) 𝑓(𝑥) =1
𝑥−5, 𝑥 ≠ 5
(c) 𝑓(𝑥) =𝑥2−25
𝑥+5, 𝑥 ≠ −5
(d) 𝑓(𝑥) = |𝑥 − 5|
Solution:
(a) Given function 𝑓(𝑥) = 𝑥 − 5
Let 𝑘 be any real number. At 𝑥 = 𝑘, 𝑓(𝑘) = 𝑘 − 5
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
(𝑥 − 5) = 𝑘 − 5
RHL = lim𝑥→𝑘+
𝑓(𝑥) = lim𝑥→𝑘+
(𝑥 − 5) = 𝑘 − 5
At 𝑥 = 𝑘, LHL = RHL = 𝑓(𝑘) = 𝑘 − 5
Hence, the function 𝑓 is continuous for all real numbers.
(b) Given function 𝑓(𝑥) =1
𝑥−5, 𝑥 ≠ 5
Let 𝑘(𝑘 ≠ 5) be any real number. At 𝑥 = 𝑘, 𝑓(𝑘) =1
𝑘−5
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
(1
𝑥−5) =
1
𝑘−5
RHL = lim𝑥→𝑘+
𝑓(𝑥) = lim𝑥→𝑘+
(1
𝑥−5) =
1
𝑘−5
At 𝑥 = 𝑘, LHL = RHL = 𝑓(𝑘) =1
𝑘−5
Hence, the function 𝑓 is continuous for all real numbers (except 5).
-
Class-XII-Maths Continuity and Differentiabil ity
3 Practice more on Continuity and Differentiability www.embibe.com
(c) Given function 𝑓(𝑥) =𝑥2−25
𝑥+5, 𝑥 ≠ −5
Let 𝑘(𝑘 ≠ −5) be any real number.
At 𝑥 = 𝑘, 𝑓(𝑘) =𝑘2−25
𝑘+5=
(𝑘+5)(𝑘−5)
(𝑘+5)= (𝑘 + 5)
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
(𝑥2−25
𝑥+5) = lim
𝑥→𝑘−((𝑘+5)(𝑘−5)
(𝑘+5)) = 𝑘 + 5
RHL = lim𝑥→𝑘+
𝑓(𝑥) = lim𝑥→𝑘+
(𝑥2−25
𝑥+5) = lim
𝑥→𝑘+((𝑘+5)(𝑘−5)
(𝑘+5)) = 𝑘 + 5
At 𝑥 = 𝑘, LHL = RHL = 𝑓(𝑘) = 𝑘 + 5
Hence, the function 𝑓 is continuous for all real numbers (except −5).
(d) Given function is 𝑓(𝑥) = |𝑥 − 5| = {5 − 𝑥, 𝑥 < 5𝑥 − 5, 𝑥 ≥ 5
Let 𝑘 be any real number. According to question,𝑘 can be 𝑘 < 5 or 𝑘 = 5 or 𝑘 > 5.
First case: If 𝑘 < 5,
𝑓(𝑘) = 5 − 𝑘 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(5 − 𝑥) =5 − 𝑘. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 5.
Second case: If 𝑘 = 5,
𝑓(𝑘) = 𝑘 − 5 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 − 5) = 𝑘 − 5. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 𝑥 = 5.
Third case: If 𝑘 > 5,
𝑓(𝑘) = 𝑘 − 5 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 − 5) = 𝑘 − 5. Here, lim𝑥→𝑘
𝑓(𝑥) =𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 5.
Hence, the function 𝑓 is continuous for all real numbers.
4. Prove that the function 𝑓(𝑥) = 𝑥𝑛, is continuous at 𝑥 = 𝑛, where 𝑛 is a positive integer.
Solution:
Given function is 𝑓(𝑥) = 𝑥𝑛.
At 𝑥 = 𝑛, 𝑓(𝑛) = 𝑥𝑛
lim𝑥→𝑛
𝑓(𝑥) = lim𝑥→𝑛
(𝑥𝑛) = 𝑥𝑛
Here, at 𝑥 = 𝑛, lim𝑥→𝑛
𝑓(𝑥) =𝑓(𝑛) = 𝑥𝑛
Since lim𝑥→𝑛
𝑓(𝑥) =𝑓(𝑛) = 𝑥𝑛
-
Class-XII-Maths Continuity and Differentiabil ity
4 Practice more on Continuity and Differentiability www.embibe.com
Hence, the function 𝑓 is continuous at 𝑥 = 𝑛, where 𝑛 is positive integer.
5. Is the function 𝑓 defined by 𝑓(𝑥) = {𝑥, 𝑥 ≤ 15, 𝑥 > 1
continuous at 𝑥 = 0? At 𝑥 = 1? At 𝑥 = 2?
Solution:
Given function is 𝑓(𝑥) = {𝑥, 𝑥 ≤ 15, 𝑥 > 1
At 𝑥 = 0, 𝑓(0) = 0
lim𝑥→0
𝑓(𝑥) = lim𝑥→0
(𝑥) = 0
Here at 𝑥 = 0, lim𝑥→0
𝑓(𝑥) = 𝑓(0) = 0
Hence, the function 𝑓 is discontinuous at 𝑥 = 0.
At 𝑥 = 1, 𝑓(1) = 1
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(𝑥) = 1
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(5) =5
Here, at 𝑥 = 1, LHL ≠ RHL.
Hence, the function 𝑓 is discontinuous at 𝑥 = 1.
At 𝑥 = 2, 𝑓(2) = 5
lim𝑥→2
𝑓(𝑥) = lim𝑥→2
(5) = 5
Here, at 𝑥 = 2, lim𝑥→2
𝑓(𝑥) = 𝑓(2) = 5
Hence, the function 𝑓 is continuous at 𝑥 = 2.
6. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {2𝑥 + 3, If 𝑥 ≤ 22𝑥 − 3, If 𝑥 > 2
Solution:
Let 𝑘 be any real number. According to question, 𝑘 < 2 or 𝑘 = 2 or 𝑘 > 2
First case: 𝑘 < 2
𝑓(𝑘) = 2𝑘 + 3 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(2𝑥 + 3) = 2𝑘 + 3. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers smaller than 2.
-
Class-XII-Maths Continuity and Differentiabil ity
5 Practice more on Continuity and Differentiability www.embibe.com
Second case: If 𝑘 = 2, 𝑓(2) = 2𝑘 + 3
LHL = lim𝑥→2−
𝑓(𝑥) = lim𝑥→2−
(2𝑥 + 3) = 7
RHL = lim𝑥→2+
𝑓(𝑥) = lim𝑥→2+
(2𝑥 − 3) = 1
Here, at 𝑥 = 2, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 2.
Third case: If 𝑘 > 2,
𝑓(𝑘) = 2𝑘 − 3 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(2𝑥 − 3) = 2𝑘 − 3. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Therefore, the function 𝑓 is continuous for all real numbers greater than 2.
Hence, the function 𝑓 is discontinuous only at 𝑥 = 2.
7. Find all points of discontinuity of 𝑓,
where 𝑓 is defined by 𝑓(𝑥) = {|𝑥| + 3, If 𝑥 ≤ −3
−2𝑥, If − 3 < 𝑥 < 36𝑥 + 2, If 𝑥 ≥ 3
Solution:
Let 𝑘 be any real number. According to question,
𝑘 < −3 or 𝑘 = −3 or −3 < 𝑘 < 3 or 𝑘 = 3 or 𝑘 > 3
First case: If 𝑘 < −3,
𝑓(𝑥) = −𝑘 + 3 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(−𝑥 + 3) = −𝑘 + 3. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than −3.
Second case: If 𝑘 = −3, 𝑓(−3) = −(−3) + 3 = 6
LHL = lim𝑥→−3−
𝑓(𝑥) = lim𝑥→−3−
(−𝑥 + 3) = −(−3) + 3 = 6
RHL = lim𝑥→−3+
𝑓(𝑥) = lim𝑥→−3+
(−2𝑥) = −2(−3) = 6. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 𝑥 = −3.
Third case: If −3 < 𝑘 < 3,
𝑓(𝑘) = −2𝑘 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
𝑓(−2𝑥) = −2𝑘. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at −3 < 𝑥 < 3.
Fourth case: If 𝑘 = 3,
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
(−2𝑥) = −2𝑘
RHL = lim𝑥→𝑘+
𝑓(𝑥) = lim𝑥→𝑘+
(6𝑥 + 2) = 6𝑘 + 2,
Here, at 𝑥 = 3, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 3.
-
Class-XII-Maths Continuity and Differentiabil ity
6 Practice more on Continuity and Differentiability www.embibe.com
Fifth case: If 𝑘 > 3,
𝑓(𝑘) = 6𝑘 + 2 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(6𝑥 + 2) = 6𝑘 + 2. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all numbers greater than 3.
Hence, the function 𝑓 is discontinuous only at 𝑥 = 3.
8. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {|𝑥|
𝑥, If 𝑥 ≠ 0
0, If 𝑥 = 0
Solution:
After redefining the function 𝑓, we get
𝑓(𝑥) =
{
−
𝑥
𝑥= −1, If 𝑥 < 0
0, If 𝑥 = 0𝑥
𝑥= 1, If 𝑥 > 0
Let 𝑘 be any real number. According to question, 𝑘 < 0 or 𝑘 = 0 or 𝑘 > 0.
First case: If 𝑘 < 0,
𝑓(𝑘) = −𝑘
𝑘= −1 and lim
𝑥→𝑘𝑓(𝑥) = lim
𝑥→𝑘(−
𝑥
𝑥) = −1. Hence, lim
𝑥→𝑘𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers smaller than 0.
Second case: If, 𝑘 = 0, 𝑓(0) = 0
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
(−𝑥
𝑥) = −1 and RHL = lim
𝑥→𝑘+𝑓(𝑥) = lim
𝑥→𝑘+(𝑥
𝑥) = 1.
Here, at 𝑥 = 0, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 0.
Third case: If 𝑘 > 0,
𝑓(𝑘) =𝑘
𝑘= 1 and lim
𝑥→𝑘𝑓(𝑥) = lim
𝑥→𝑘(𝑥
𝑥) = 1. Here, lim
𝑥→𝑘𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 0.
Therefore, the function 𝑓 is discontinuous only at 𝑥 = 0.
9. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {𝑥
|𝑥|, If 𝑥 < 0
−1, If 𝑥 ≥ 0
-
Class-XII-Maths Continuity and Differentiabil ity
7 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Redefining the function, we get
𝑓(𝑥) = {
𝑥
|𝑥|=
𝑥
−𝑥= −1, If 𝑥 < 0
−1, If 𝑥 ≥ 0
Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘) = −1, where 𝑘 is a real number.
Hence, the function 𝑓 is continuous for all real numbers.
10. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {𝑥 + 1, If ≥ 1
𝑥2 + 1, If 𝑥 < 1
Solution:
Given function is defined by 𝑓(𝑥) = {𝑥 + 1, If ≥ 1
𝑥2 + 1, If 𝑥 < 1
Let 𝑘 be any real number. According to question, 𝑘 < 1 or 𝑘 = 1 or 𝑘 > 1
First case: If 𝑘 < 1,
𝑓(𝑘) = 𝑘2 + 1 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥2 + 1) = 𝑘2 + 1. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers smaller than 1.
Second case: If 𝑘 = 1, 𝑓(1) = 1 + 1 = 2
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(𝑥2 + 1) = 1 + 1 = 2
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(𝑥 + 1) = 1 + 1 = 2,
Here, at 𝑥 = 1, LHL = RHL = 𝑓(1). Hence, the function 𝑓 is continuous at 𝑥 = 1.
Third case: If 𝑘 > 1,
𝑓(𝑘) = 𝑘 + 1 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 + 1) = 𝑘 + 1 . Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 1.
Therefore, the function 𝑓 is continuous for all real numbers.
11. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {𝑥3 − 3, If 𝑥 ≤ 2
𝑥2 + 1, If 𝑥 > 2
-
Class-XII-Maths Continuity and Differentiabil ity
8 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given function is defined by 𝑓(𝑥) = {𝑥3 − 3, If 𝑥 ≤ 2
𝑥2 + 1, If 𝑥 > 2
Let 𝑘 be any real number. According to question, 𝑘 < 2 or 𝑘 = 2 or 𝑘 > 2
First case: If 𝑘 < 2,
𝑓(𝑘) = 𝑘3 − 3 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥3 − 3) = 𝑘3 − 3. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 2.
Second case: If 𝑘 = 2, 𝑓(2) = 23 − 3 = 5
LHL = lim𝑥→2−
𝑓(𝑥) = lim𝑥→2−
(𝑥3 − 3) = 23 − 3 = 5
RHL = lim𝑥→2+
𝑓(𝑥) = lim𝑥→2+
(𝑥2 + 1) = 22 + 1 = 5
Here at 𝑥 = 2, LHL = RHL = 𝑓(2)
Hence, the function 𝑓 is continuous at 𝑥 = 2.
Third case: If 𝑘 > 2,
𝑓(𝑘) = 𝑘2 + 1 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥2 + 1) = 𝑘2 + 1. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for real numbers greater than 2.
Hence, the function 𝑓 is continuous for all real numbers.
12. Find all points of discontinuity of 𝑓, where 𝑓 is defined by 𝑓(𝑥) = {𝑥10 − 1, If 𝑥 ≤ 1
𝑥2, If 𝑥 > 1
Solution:
Given function is defined by 𝑓(𝑥) = {𝑥10 − 1, If 𝑥 ≤ 1
𝑥2, If 𝑥 > 1
Let 𝑘 be any real number. According to question, 𝑘 < 1 or 𝑘 = 1 or 𝑘 > 1
First case: If 𝑘 < 1,
𝑓(𝑘) = 𝑘10 − 1 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥10 − 1) = 𝑘10 − 1. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 1.
Second case: If 𝑘 = 1, 𝑓(1) = 110 − 1 = 0
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(𝑥10 − 1) = 0
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(𝑥2) = 1
Here at 𝑥 = 1, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 1.
-
Class-XII-Maths Continuity and Differentiabil ity
9 Practice more on Continuity and Differentiability www.embibe.com
Third case: If 𝑘 > 1,
𝑓(𝑘) = 𝑘2 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥2) = 𝑘2. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real values greater than 1.
Hence, the function 𝑓 is discontinuous only at 𝑥 = 1.
13. Is the function defined by 𝑓(𝑥) = {𝑥 + 5, If 𝑥 ≤ 1 𝑥 − 5, If 𝑥 > 1
a continuous function?
Solution:
Given function is defined by 𝑓(𝑥) = {𝑥 + 5, If 𝑥 ≤ 1 𝑥 − 5, If 𝑥 > 1
Let, 𝑘 be any real number. According to question, 𝑘 < 1 or 𝑘 = 1 or 𝑘 > 1
First case: If 𝑘 < 1,
𝑓(𝑘) = 𝑘 + 5 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 + 5) = 𝑘 + 5. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 1.
Second case: If 𝑘 = 1, 𝑓(1) = 1 + 5 = 6
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(𝑥 + 5) = 6
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(𝑥 − 5) = −4,
Here at 𝑥 = 1, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 1.
Third case: If 𝑘 > 1,
𝑓(𝑘) = 𝑘 − 5 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 − 5) = 𝑘 − 5
Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 1.
Therefore, the function 𝑓 is discontinuous only at 𝑥 = 1.
14. Discuss the continuity of the function 𝑓,
where 𝑓 is defined by 𝑓(𝑥) = { 3, If 0 ≤ 𝑥 ≤ 14, If 1 < 𝑥 < 3
5, If 3 ≤ 𝑥 ≤ 10
-
Class-XII-Maths Continuity and Differentiabil ity
10 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given function is defined by 𝑓(𝑥) = { 3, If 0 ≤ 𝑥 ≤ 14, If 1 < 𝑥 < 3
5, If 3 ≤ 𝑥 ≤ 10
Let 𝑘 be any real number. According to question, 𝑘 can be
0 ≤ 𝑘 ≤ 1 or 𝑘 = 1 or 1 < 𝑘 < 3 or 𝑘 = 3 or 3 ≤ 𝑘 ≤ 10
First case: If 0 ≤ 𝑘 ≤ 1,
𝑓(𝑘) = 3 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(3) = 3. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for 0 ≤ 𝑥 ≤ 1.
Second case: If 𝑘 = 1, 𝑓(1) = 3
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(3) = 3
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(4) = 4,
Here at 𝑥 = 1, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 1.
Third case: If 1 < 𝑘 < 3,
𝑓(𝑘) = 4 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(4) = 4. Here lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for 1 < 𝑥 < 3.
Fourth case: If 𝑘 = 3,
LHL = lim𝑥→3−
𝑓(𝑥) = lim𝑥→3−
(4) = 4 and RHL = lim𝑥→3+
𝑓(𝑥) = lim𝑥→3+
(5) = 5,
Here at 𝑥 = 3, LHL ≠ RHL. Hence, the function 𝑓 is discontinuous at 𝑥 = 3.
Fifth case: If 3 ≤ 𝑘 ≤ 10,
𝑓(𝑘) = 5 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(5) = 5. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for 3 ≤ 𝑥 ≤ 10.
Hence, the function 𝑓 is discontinuous only at 𝑥 = 1 and 𝑥 = 3.
15. Discuss the continuity of the function 𝑓,
where 𝑓 is defined by 𝑓(𝑥) = {2𝑥, If 𝑥 < 00, If 0 ≤ 𝑥 ≤ 1 4𝑥, If 𝑥 > 1
Solution:
Given function is defined by 𝑓(𝑥) = {2𝑥, If 𝑥 < 00, If 0 ≤ 𝑥 ≤ 1 4𝑥, If 𝑥 > 1
-
Class-XII-Maths Continuity and Differentiabil ity
11 Practice more on Continuity and Differentiability www.embibe.com
Let 𝑘 be any real number. According to question,
𝑘 < 0 or 𝑘 = 0 or 0 ≤ 𝑘 ≤ 1 or 𝑘 = 1 or 𝑘 > 1
First case: If 𝑘 < 0,
𝑓(𝑘) = 2𝑘 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(2𝑥) = 2𝑘. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 0.
Second case: If 𝑘 = 0, 𝑓(0) = 0
LHL = lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
(2𝑥) = 0
RHL = lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
(0) = 0. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 𝑥 = 0.
Third case: If 0 ≤ 𝑘 ≤ 1,
𝑓(𝑘) = 0 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(0) = 0. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 0 ≤ 𝑥 ≤ 1.
Fourth case: If 𝑘 = 1,
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(0) = 0
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(4𝑥) = 4,
Here, at 𝑥 = 1, LHL ≠ RHL.
Hence, the function 𝑓 is discontinuous at 𝑥 = 1.
Fifth case: If 𝑘 > 1,
𝑓(𝑘) = 4𝑘 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(4𝑥) = 4𝑘.
Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 1.
Hence, the function 𝑓 is discontinuous only at 𝑥 = 1.
16. Discuss the continuity of the function 𝑓,
where 𝑓 is defined by 𝑓(𝑥) = {−2, If 𝑥 ≤ −12𝑥, If − 1 < 𝑥 ≤ 1 2, If 𝑥 > 1
Solution:
Given function is defined by 𝑓(𝑥) = {−2, If 𝑥 ≤ −12𝑥, If − 1 < 𝑥 ≤ 1 2, If 𝑥 > 1
-
Class-XII-Maths Continuity and Differentiabil ity
12 Practice more on Continuity and Differentiability www.embibe.com
Let 𝑘 be any real number
According to question, 𝑘 < −1 or 𝑘 = −1 or −1 < 𝑥 ≤ 1 or 𝑘 = 1 or 𝑘 > 1
First case: If 𝑘 < −1,
𝑓(𝑘) = −2 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(−2) = −2. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than −1.
Second case: If 𝑘 = −1, 𝑓(−1) = −2
LHS = lim𝑥→−1−
𝑓(𝑥) = lim𝑥→−1−
(−2) = −2
RHL = lim𝑥→−1+
𝑓(𝑥) = lim𝑥→−1+
(2𝑥) = −2. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 𝑥 = −1.
Third case: If −1 < 𝑥 ≤ 1,
𝑓(𝑘) = 2𝑘 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(2𝑥) = 2𝑘. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at −1 < 𝑥 ≤ 1.
Fourth case: If 𝑘 = 1,
LHL = lim𝑥→1−
𝑓(𝑥) = lim𝑥→1−
(2𝑥) = 2
RHL = lim𝑥→1+
𝑓(𝑥) = lim𝑥→1+
(2) = 2. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous at 𝑥 = 1.
Fifth case: If 𝑘 > 1,
𝑓(𝑘) = 2 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(2) = 2.
Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 1.
Therefore, the function 𝑓 is continuous for all real numbers.
17. Find the relationship between 𝑎 and 𝑏 so that the function 𝑓 defined by
𝑓(𝑥) = {𝑎𝑥 + 1, If 𝑥 ≤ 3
𝑏𝑥 + 3, If 𝑥 > 3 is continuous at 𝑥 = 3.
Solution:
Given functions is defined by 𝑓(𝑥) = {𝑎𝑥 + 1, If 𝑥 ≤ 3
𝑏𝑥 + 3, If 𝑥 > 3
Given that the function is continuous at 𝑥 = 3. Therefore, LHL = RHL = 𝑓(3)
⇒ lim𝑥→3−
𝑓(𝑥) = lim𝑥→3+
𝑓(𝑥) = 𝑓(3)
-
Class-XII-Maths Continuity and Differentiabil ity
13 Practice more on Continuity and Differentiability www.embibe.com
⇒ lim𝑥→3−
𝑎𝑥 + 1 = lim𝑥→3+
𝑏𝑥 + 3 = 3𝑎 + 1
⇒ 3𝑎 + 1 = 3𝑏 + 3 = 3𝑎 + 1
⇒ 3𝑎 = 3𝑏 + 2 ⇒ 𝑎 = 𝑏 +2
3
Hence, the relationship between 𝑎 and 𝑏 is 𝑎 = 𝑏 +2
3
18. For what value of 𝜆 is the function defined by
𝑓(𝑥) = {𝜆(𝑥2 − 2𝑥), If 𝑥 ≤ 0
4𝑥 + 1, If 𝑥 > 0
Continuous at 𝑥 = 0? What about continuity at 𝑥 = 1?
Solution:
Given function is defined as 𝑓(𝑥) = {𝜆(𝑥2 − 2𝑥), If 𝑥 ≤ 0
4𝑥 + 1, If 𝑥 > 0
Given that the function is continuous at 𝑥 = 0. Therefore, LHL = RHL = 𝑓(0)
⇒ lim𝑥→0−
𝑓(𝑥) = lim𝑥→0+
𝑓(𝑥) = 𝑓(0)
⇒ lim𝑥→0−
𝜆(𝑥2 − 2𝑥) = lim𝑥→0+
4𝑥 + 1 = 𝜆[(0)2 − 2(0)]
⇒ 𝜆[(0)2 − 2(0)] = 4(0) + 1 = 𝜆(0)
⇒ 0. 𝜆 = 1 ⇒ 𝜆 =1
0
Hence, there is no real value of 𝜆 for which the given function be continuous.
If 𝑥 = 1,
𝑓(1) = 4(1) + 1 = 5 and lim𝑥→1
𝑓(𝑥) = lim𝑥→1
4(1) + 1 = 5. Here, lim𝑥→1
𝑓(𝑥) = 𝑓(1)
Therefore, the function 𝑓 is continuous for all real values of 𝜆.
19. Show that the function defined by 𝑔(𝑥) = 𝑥 − [𝑥] is discontinuous at all integral points.
Here [𝑥] denotes the greatest integer less than or equal to 𝑥.
Solution:
Given function is defined by 𝑔(𝑥) = 𝑥 − [𝑥]
Let 𝑘 be any integer
LHL = lim𝑥→𝑘−
𝑓(𝑥) = lim𝑥→𝑘−
𝑥 − [𝑥] = 𝑘 − (𝑘 − 1) = 1
-
Class-XII-Maths Continuity and Differentiabil ity
14 Practice more on Continuity and Differentiability www.embibe.com
RHL = lim𝑥→𝑘+
𝑓(𝑥) = lim𝑥→𝑘+
𝑥 − [𝑥] = 𝑘 − (𝑘) = 0,
Here, at 𝑥 = 𝑘, LHL ≠ RHL.
Therefore, the function 𝑓 is discontinuous for all integers.
20. Is the function defined by 𝑓(𝑥) = 𝑥2 − sin𝑥 + 5 continuous at 𝑥 = 𝜋.
Solution:
Given function is defined by 𝑓(𝑥) = 𝑥2 − sin𝑥 + 5,
At 𝑥 = 𝜋, 𝑓(𝜋) = 𝜋2 − sin𝜋 + 5 = 𝜋2 − 0 + 5 = 𝜋2 + 5
lim𝑥→𝑛
𝑓(𝑥) = lim𝑥→𝑛
𝑥2 − sin𝑥 + 5 = 𝜋2 − sin𝜋 + 5 = 𝜋2 − 0 + 5 = 𝜋2 + 5
Here, at 𝑥 = 𝜋, lim𝑥→𝑛
𝑓(𝑥) = 𝑓(𝜋) = 𝜋2 + 5
Therefore, the function 𝑓(𝑥) is continuous at 𝑥 = 𝜋.
21. Discuss the continuity of the following functions:
(a) 𝑓(𝑥) = sin 𝑥 + cos 𝑥
(b) 𝑓(𝑥) = sin 𝑥 − cos𝑥
(c) 𝑓(𝑥) = sin𝑥. cos 𝑥
Solution:
Let 𝑔(𝑥) = sin 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, 𝑔(𝑘) = sin𝑘
LHL = lim𝑥→𝑘−
𝑔(𝑥) = lim𝑥→𝑘−
sin 𝑥 = limℎ→0
sin(𝑘 − ℎ) = limℎ→0
sin𝑘 cosℎ − cos 𝑘 sin ℎ = sin 𝑘
RHL = lim𝑥→𝑘+
𝑔(𝑥) = lim𝑥→𝑘+
sin 𝑥 = limℎ→0
sin(𝑘 + ℎ) = limℎ→0
sin 𝑘 cos ℎ + cos 𝑘 sinℎ = sin 𝑘
Here, at 𝑥 = 𝑘, LHL = RHL = 𝑔(𝑘).
Hence, the function 𝑔 is continuous for all real numbers.
Let ℎ(𝑥) = cos 𝑥
Let 𝑘 be any real number. 𝑥 = 𝑘, ℎ(𝑘) = cos 𝑘
𝐿𝐻𝐿 = lim𝑥→𝑘−
ℎ(𝑥) = lim𝑥→𝑘−
cos 𝑥 = limℎ→0
cos(𝑘 − ℎ) = limℎ→0
cos𝑘 cos ℎ + sin𝑘 sinℎ = cos 𝑘
RHL = lim𝑥→𝑘+
ℎ(𝑥) = lim𝑥→𝑘+
cos 𝑥 = limℎ→0
cos(𝑘 + ℎ) = limℎ→0
cos 𝑘 cosℎ − sin𝑘 sinℎ = cos 𝑘
-
Class-XII-Maths Continuity and Differentiabil ity
15 Practice more on Continuity and Differentiability www.embibe.com
Here, at 𝑥 = 𝑘, LHL = RHL = ℎ(𝑘).
Hence, the function ℎ is continuous for all real numbers.
We know that if 𝑔 and ℎ are two continuous functions, then the functions 𝑔 + ℎ, 𝑔 − ℎ and 𝑔ℎ also be a continuous function.
Therefore, (a) 𝑓(𝑥) = sin𝑥 + cos 𝑥 (b) 𝑓(𝑥) = sin 𝑥 − cos𝑥 and
(c) 𝑓(𝑥) = sin𝑥. cos 𝑥 are continuous functions.
22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
Let 𝑔(𝑥) = sin 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, 𝑔(𝑘) = sin𝑘
LHL = lim𝑥→𝑘−
𝑔(𝑥) = lim𝑥→𝑘−
sin 𝑥 = limℎ→0
sin(𝑘 − ℎ) = limℎ→0
sin𝑘 cos ℎ − cos 𝑘 sin ℎ = sin𝑘
RHL = lim𝑥→𝑘+
𝑔(𝑥) = lim𝑥→𝑘+
sin 𝑥 = limℎ→0
sin(𝑘 + ℎ) = limℎ→0
sin𝑘 cosℎ + cos 𝑘 sin ℎ = sin 𝑘
Here, at 𝑥 = 𝑘, LHL = RHL = 𝑔(𝑘).
Hence, the function 𝑔 is continuous for all real numbers.
Let ℎ(𝑥) = cos 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, ℎ(𝑘) = cos𝑘
𝐿𝐻𝐿 = lim𝑥→𝑘−
ℎ(𝑥) = lim𝑥→𝑘−
cos 𝑥 = limℎ→0
cos(𝑘 − ℎ) = limℎ→0
cos 𝑘 cosℎ + sin𝑘 sinℎ = cos 𝑘
RHL = lim𝑥→𝑘+
ℎ(𝑥) = lim𝑥→𝑘+
cos 𝑥 = limℎ→0
cos(𝑘 + ℎ) = limℎ→0
cos 𝑘 cosℎ − sin𝑘 sinℎ = cos 𝑘
Here, at 𝑥 = 𝑘, LHL = RHL = ℎ(𝑘).
Hence, the function ℎ is continuous for all real numbers.
We know that if 𝑔 and ℎ are two continuous functions, then the functions 𝑔
ℎ, ℎ ≠ 0,
1
ℎ, ℎ ≠ 0
and 1
𝑔, 𝑔 ≠ 0 are continuous functions.
Therefore, cosec𝑥 =1
sin𝑥, sin 𝑥 ≠ 0 is continuous ⇒ 𝑥 ≠ 𝑛𝜋(𝑛 ∈ 𝑍) is continuous.
Hence, cosec 𝑥 is continuous except 𝑥 = 𝑛𝜋(𝑛 ∈ 𝑍).
sec 𝑥 =1
cos𝑥, cos 𝑥 ≠ 0 is continuous. ⇒ 𝑥 ≠
(2𝑛+1)𝜋
2(𝑛 ∈ 𝑍) is continuous.
Hence, sec 𝑥 is continuous except 𝑥 =(2𝑛+1)𝜋
2(𝑛 ∈ 𝑍).
cot 𝑥 =cos𝑥
sin𝑥, sin 𝑥 ≠ 0 is continuous. ⇒ 𝑥 ≠ 𝑛𝜋(𝑛 ∈ 𝑍) is continuous.
-
Class-XII-Maths Continuity and Differentiabil ity
16 Practice more on Continuity and Differentiability www.embibe.com
Hence, cot 𝑥 is continuous except 𝑥 = 𝑛𝜋(𝑛 ∈ 𝑍).
23. Find all points of discontinuity of 𝑓, where 𝑓(𝑥) = {sin𝑥
𝑥, If 𝑥 < 0
𝑥 + 1, If 𝑥 ≥ 0
Solution:
Given function is defined by 𝑓(𝑥) = {sin𝑥
𝑥, If 𝑥 < 0
𝑥 + 1, If 𝑥 ≥ 0
Let 𝑘 be any real number. According to question, 𝑘 < 0 or 𝑘 = 0 or 𝑘 > 0
First case: If 𝑘 < 0
𝑓(𝑘) =sin𝑘
𝑘 and lim
𝑥→𝑘𝑓(𝑥) = lim
𝑥→𝑘(sin𝑥
𝑥) =
sin𝑘
𝑘. Here, lim
𝑥→𝑘𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers less than 0.
Second case: If 𝑘 = 0
𝑓(0) = 0 + 1 = 1
LHL = lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
(𝑥 + 1) = 0 + 1 = 1
RHL = lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
(𝑥 + 1) = 0 + 1 = 1,
Here at 𝑥 = 0, LHL = RHL = 𝑓(0). Hence, the function 𝑓 is continuous at 𝑥 = 0.
Third case: If 𝑘 > 0
𝑓(𝑘) = 𝑘 + 1 and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(𝑥 + 1) = 𝑘 + 1. Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for all real numbers greater than 0.
Therefore, the function 𝑓 is continuous for all real numbers.
24. Determine if 𝑓 defined by
𝑓(𝑥) = {𝑥2 sin
1
𝑥, If 𝑥 ≠ 0
0, If 𝑥 = 0
is a continuous function?
Solution:
Given function is defined by 𝑓(𝑥) = {𝑥2 sin
1
𝑥, If 𝑥 ≠ 0
0, If 𝑥 = 0
-
Class-XII-Maths Continuity and Differentiabil ity
17 Practice more on Continuity and Differentiability www.embibe.com
Let 𝑘 be any real number. According to question, 𝑘 ≠ 0 or 𝑘 = 0
First case: If 𝑘 ≠ 0
𝑓(𝑘) = 𝑘2 sin1
𝑘 and lim
𝑥→𝑘𝑓(𝑥) = lim
𝑥→𝑘(𝑥2 sin
1
𝑥) = 𝑘2 sin
1
𝑘. Here, lim
𝑥→𝑘𝑓(𝑥) = 𝑓(𝑘)
Hence, the function 𝑓 is continuous for 𝑘 ≠ 0.
Second case: If, 𝑘 = 0, 𝑓(0) = 0
LHL = lim𝑥→0−
𝑓(𝑥) = lim𝑥→0−
(𝑥2 sin1
𝑥) = lim
𝑥→0(𝑥2 sin
1
𝑥)
We know that, −1 ≤ sin1
𝑥≤ 1, 𝑥 ≠ 0 ⇒ −𝑥2 ≤ sin
1
𝑥≤ 𝑥2
⇒ lim𝑥→0
(−𝑥2) ≤ lim𝑥→0
sin1
𝑥≤ lim𝑥→0
𝑥2
⇒ 0 ≤ lim𝑥→0
sin1
𝑥≤ 0 ⇒ lim
𝑥→0sin
1
𝑥= 0 ⇒ lim
𝑥→0−𝑥2 sin
1
𝑥= 0 ⇒ lim
𝑥→0−𝑓(𝑥) = 0
Similarly, RHL = lim𝑥→0+
𝑓(𝑥) = lim𝑥→0+
(𝑥2 sin1
𝑥) = lim
𝑥→0(𝑥2 sin
1
𝑥) = 0
Here, at 𝑥 = 0, LHL = RHL = 𝑓(0)
Hence, at 𝑥 = 0, 𝑓 is continuous.
Therefore, the function 𝑓 is continuous for all real numbers.
25. Examine the continuity of 𝑓, where 𝑓 is defined by
𝑓(𝑥) = {sin 𝑥 − cos𝑥 , If 𝑥 ≠ 0
−1, If 𝑥 = 0
Solution:
Given function is defined by 𝑓(𝑥) = {sin𝑥 − cos 𝑥 , If 𝑥 ≠ 0
−1, If 𝑥 = 0
Let 𝑘 be any real number. According to question, 𝑘 ≠ 0 or 𝑘 = 0
First case: If 𝑘 ≠ 0, 𝑓(0) = 0 − 1 = −1
LHL = lim𝑘→0−
𝑓(𝑥) = lim𝑘→0−
(sin 𝑥 − cos 𝑥) = 0 − 1 = −1
RHL = lim𝑘→0+
𝑓(𝑥) = lim𝑘→0+
(sin 𝑥 − cos 𝑥) = 0 − 1 = −1
Hence, at 𝑥 ≠ 0, LHL = RHL = 𝑓(𝑥)
Hence, the function 𝑓 is continuous at 𝑥 ≠ 0.
Second case: If, 𝑘 = 0, 𝑓(𝑘) = −1
and lim𝑥→𝑘
𝑓(𝑥) = lim𝑥→𝑘
(−1) = −1 Here, lim𝑥→𝑘
𝑓(𝑥) = 𝑓(𝑘)
-
Class-XII-Maths Continuity and Differentiabil ity
18 Practice more on Continuity and Differentiability www.embibe.com
Hence, the function 𝑓 is continuous at 𝑥 = 0
Hence, the function 𝑓 is continuous for all real numbers.
26. Find the values of 𝑘 so that the function 𝑓 is continuous at the indicated point.
𝑓(𝑥) = {
𝑘 cos𝑥
𝜋−2𝑥, If 𝑥 ≠
𝜋
2
3, If 𝑥 =𝜋
2
at 𝑥 =𝜋
2
Solution:
Given function is defined by 𝑓(𝑥) = {
𝑘 cos𝑥
𝜋−2𝑥, If 𝑥 ≠
𝜋
2
3, If 𝑥 =𝜋
2
at 𝑥 =𝜋
2
Given that the function is continuous at 𝑥 =𝜋
2. Therefore, LHL = RHL = 𝑓 (
𝜋
2)
⇒ lim𝑥→
𝜋2
𝑓(𝑥) = lim𝑥→
𝜋+
2
𝑓(𝑥) = 𝑓 (𝜋
2)
⇒ lim𝑥→
𝜋2
𝑘 cos 𝑥
𝜋 − 2𝑥= lim
𝑥→𝜋+
2
𝑘 cos 𝑥
𝜋 − 2𝑥= 3
⇒ limℎ→0
𝑘 cos(𝜋
2−ℎ)
𝜋−2(𝜋
2−ℎ)
= limℎ→0
𝑘 cos(𝜋
2+ℎ)
𝜋−2(𝜋
2+ℎ)
= 3
⇒ limℎ→0
𝑘 sinℎ
2ℎ= lim
ℎ→0
−𝑘 sinℎ
−2ℎ= 3
⇒𝑘
2=
𝑘
2= 3 [∵ lim
ℎ→0
sinℎ
ℎ= 1]
⇒ 𝑘 = 6
Hence, for 𝑘 = 6, the given function 𝑓 is continuous at the indicated point.
27. Find the values of 𝑘 so that the function 𝑓 is continuous at the indicated point.
𝑓(𝑥) = {𝑘𝑥2, If 𝑥 ≤ 2
3, If 𝑥 > 2 at 𝑥 = 2
Solution:
Given function is defined by 𝑓(𝑥) = {𝑘𝑥2, If 𝑥 ≤ 2
3, If 𝑥 > 2 at 𝑥 = 2
-
Class-XII-Maths Continuity and Differentiabil ity
19 Practice more on Continuity and Differentiability www.embibe.com
Given that the function is continuous at 𝑥 = 2.
Therefore, LHL = RHL = 𝑓(2)
⇒ lim𝑥→2−
𝑓(𝑥) = lim𝑥→2+
𝑓(𝑥) = 𝑓(2)
⇒ lim𝑥→2−
𝑘𝑥2 = lim𝑥→2+
3 = 𝑘(2)2
⇒ 4𝑘 = 3 = 4𝑘
⇒ 𝑘 =3
4
Hence, for 𝑘 =3
4, the given function 𝑓 is continuous at the indicated point.
28. Find the values of 𝑘 so that the function 𝑓 is continuous at the indicated point.
𝑓(𝑥) = {𝑘𝑥 + 1, If 𝑥 ≤ 𝜋
cos𝑥, If 𝑥 > 𝜋 at 𝑥 = 𝜋
Solution:
Given function is 𝑓(𝑥) = {𝑘𝑥 + 1, If 𝑥 ≤ 𝜋
cos 𝑥, If 𝑥 > 𝜋 at 𝑥 = 𝜋
Given that the function is continuous at 𝑥 = 𝜋,
Therefore, LHL = RHL = 𝑓(𝜋)
⇒ lim𝑥→𝜋−
𝑓(𝑥) = lim𝑥→𝜋+
𝑓(𝑥) = 𝑓(𝜋)
⇒ lim𝑥→𝜋−
𝑘𝑥 + 1 = lim𝑥→𝜋+
cos 𝑥 = 𝑘(𝜋) + 1
⇒ 𝑘(𝜋) + 1 = cos 𝜋 = 𝑘𝜋 + 1
⇒ 𝑘𝜋 + 1 = −1 = 𝑘𝜋 + 1
⇒ 𝜋𝑘 = −2
⇒ 𝑘 = −2
𝜋
Hence, for 𝑘 = −2
𝜋, the given function 𝑓 is continuous at the indicated point.
29. Find the values of 𝑘 so that the function 𝑓 is continuous at the indicated point.
𝑓(𝑥) = {𝑘𝑥 + 1, If 𝑥 ≤ 53𝑥 − 5, If 𝑥 > 5
at 𝑥 = 5
-
Class-XII-Maths Continuity and Differentiabil ity
20 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given function is defined by 𝑓(𝑥) = {𝑘𝑥 + 1, If 𝑥 ≤ 53𝑥 − 5, If 𝑥 > 5
at 𝑥 = 5
Given that the function is continuous at 𝑥 = 5.
Therefore, LHL = RHL = 𝑓(5)
⇒ lim𝑥→5−
𝑓(𝑥) = lim𝑥→5+
𝑓(𝑥) = 𝑓(5)
⇒ lim𝑥→5−
𝑘𝑥 + 1 = lim𝑥→5+
3𝑥 − 5 = 5𝑘 + 1
⇒ 5𝑘 + 1 = 15 − 5 = 5𝑘 + 1
⇒ 5𝑘 = 9
⇒ 𝑘 =9
5
Hence, for 𝑘 =9
5, the given function 𝑓 is continuous at the indicated point.
30. Find the values of 𝑎 and 𝑏 such that the function defined by
𝑓(𝑥) = {5, If 𝑥 ≤ 2
𝑎𝑥 + 𝑏, If 2 < 𝑥 < 1021, If 𝑥 ≥ 10
is continuous function.
Solution:
Given function is 𝑓(𝑥) = {5, If 𝑥 ≤ 2
𝑎𝑥 + 𝑏, If 2 < 𝑥 < 1021, If 𝑥 ≥ 10
Given that the function is continuous at 𝑥 = 2. Therefore, LHL = RHL = 𝑓(2)
⇒ lim𝑥→2−
𝑓(𝑥) = lim𝑥→2+
𝑓(𝑥) = 𝑓(2)
⇒ lim𝑥→2−
5 = lim𝑥→2+
𝑎𝑥 + 𝑏 = 5
⇒ 2𝑎 + 𝑏 = 5 …(i)
Given that the function is continuous at 𝑥 = 10. Therefore, LHL = RHL = 𝑓(10)
⇒ lim𝑥→10−
𝑓(𝑥) = lim𝑥→10+
𝑓(𝑥) = 𝑓(10)
⇒ lim𝑥→10−
𝑎𝑥 + 𝑏 = lim𝑥→10+
21 = 21
⇒ 10𝑎 + 𝑏 = 21 …(ii)
Solving the equation (i) and (ii), we get
𝑎 = 2 𝑏 = 1
-
Class-XII-Maths Continuity and Differentiabil ity
21 Practice more on Continuity and Differentiability www.embibe.com
31. Show that the function defined by 𝑓(𝑥) = cos(𝑥2) is a continuous function.
Solution:
Given function is defined by 𝑓(𝑥) = cos(𝑥2)
Assuming that the functions are well defined for all real numbers, we can write the given function 𝑓 in the combination of 𝑔 and ℎ(𝑓 = 𝑔𝑜ℎ). Where, 𝑔(𝑥) = cos𝑥 and ℎ(𝑥) = 𝑥2, if 𝑔 and ℎ both are continuous function then 𝑓 also be continuous.
[∵ 𝑔𝑜ℎ(𝑥) = 𝑔(ℎ(𝑥)) = 𝑔(𝑥2) = cos(𝑥2)]
Let the function 𝑔(𝑥) be cos 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, 𝑔(𝑘) = cos 𝑘
lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
cos 𝑥 = limℎ→0
cos(𝑘 + ℎ) = limℎ→0
cos 𝑘 cos ℎ − sin𝑘 sin ℎ = cos𝑘
Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘). Hence, t\he function 𝑔 is continuous for all real numbers.
And let the function ℎ(𝑥) be 𝑥2
Let 𝑘 be any real number. At 𝑥 = 𝑘, ℎ(𝑘) = 𝑘2
lim𝑥→𝑘
ℎ(𝑥) = lim𝑥→𝑘
𝑥2 = 𝑘2
Here, lim𝑥→𝑘
ℎ(𝑥) = ℎ(𝑘). Hence, the function ℎ is continuous for all real numbers.
Therefore, 𝑔 and ℎ both are continuous function. Hence, 𝑓 is continuous.
32. Show that the function defined by 𝑓(𝑥) = |cos 𝑥| is a continuous function.
Solution:
Given that the function is defined by 𝑓(𝑥) = |cos 𝑥|
Assuming that the functions are well defined for all real numbers, we can write the given function 𝑓 in the combination of 𝑔 and ℎ(𝑓 = 𝑔𝑜ℎ). Where, 𝑔(𝑥) = |𝑥| and ℎ(𝑥) = cos 𝑥. If 𝑔 and ℎ both are continuous function then 𝑓 also be continuous.
[∵ 𝑔𝑜ℎ(𝑥) = 𝑔(ℎ(𝑥)) = 𝑔(cos 𝑥) = |cos 𝑥|]
Let the function 𝑔(𝑥) be |𝑥|
Rearranging the function 𝑔, we get
𝑔(𝑥) = {−𝑥, If 𝑥 < 0𝑥, If 𝑥 ≥ 0
Let 𝑘 be any real number. According to question, 𝑘 < 0 or 𝑘 = 0 or 𝑘 > 0
-
Class-XII-Maths Continuity and Differentiabil ity
22 Practice more on Continuity and Differentiability www.embibe.com
First case: If 𝑘 < 0,
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(−𝑥) = 0, here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers less than 0.
Second case: If 𝑘 = 0, 𝑔(0) = 0 + 1 = 1
LHL = lim𝑥→0−
𝑔(𝑥) = lim𝑥→0−
(−𝑥) = 0
RHL = lim𝑥→0+
𝑔(𝑥) = lim𝑥→0+
(𝑥) = 0,
Here at 𝑥 = 0, LHL = RHL = 𝑔(0)
Hence, the function 𝑔 is continuous at 𝑥 = 0.
Third case: If 𝑘 > 0,
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(𝑥) = 0. Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers greater than 0.
Hence, the function 𝑔 is continuous for all real numbers.
And let the function ℎ(𝑥) be cos 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, ℎ(𝑘) = cos𝑘
lim𝑥→𝑘
ℎ(𝑥) = lim𝑥→𝑘
cos 𝑥 = cos 𝑘
Here, lim𝑥→𝑘
ℎ(𝑥) = ℎ(𝑘). Hence, the function ℎ is continuous for all real numbers.
Therefore, 𝑔 and ℎ both are continuous function. Hence, 𝑓 is continuous.
33. Examine that sin|𝑥| is a continuous function.
Solution:
Let the given function be 𝑓(𝑥) = sin|𝑥|
Assuming that the functions are well defined for all real numbers, we can write the given function 𝑓 in the combination of 𝑔 and ℎ(𝑓 = ℎ𝑜𝑔). Where, ℎ(𝑥) = sin 𝑥 and 𝑔(𝑥) = |𝑥|. If 𝑔 and ℎ both are continuous function then 𝑓 also be continuous.
[∵ ℎ𝑜𝑔(𝑥) = ℎ(𝑔(𝑥)) = ℎ(|𝑥|) = sin|𝑥|]
Function ℎ(𝑥) = sin 𝑥
Let 𝑘 be any real number. At 𝑥 = 𝑘, ℎ(𝑘) = sin𝑘
lim𝑥→𝑘
ℎ(𝑥) = lim𝑥→𝑘
sin 𝑥 = sin𝑘
Here, lim𝑥→𝑘
ℎ(𝑥) = ℎ(𝑘). Hence, the function ℎ is continuous for all real numbers.
-
Class-XII-Maths Continuity and Differentiabil ity
23 Practice more on Continuity and Differentiability www.embibe.com
Function 𝑔(𝑥) = |𝑥|
Redefining the function 𝑔, we get
𝑔(𝑥) = {−𝑥, If 𝑥 < 0𝑥, If 𝑥 ≥ 0
Let 𝑘 be any real number. According to question, 𝑘 < 0 or 𝑘 = 0 or 𝑘 > 0
First case: If 𝑘 < 0,
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(−𝑥) = 0. Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers less than 0.
Second case: If 𝑘 = 0, 𝑔(0) = 0 + 1 = 1
LHL = lim𝑥→0−
𝑔(𝑥) = lim𝑥→0−
(−𝑥) = 0
RHL = lim𝑥→0+
𝑔(𝑥) = lim𝑥→0+
(𝑥) = 0
Here at 𝑥 = 0, LHL = RHL = 𝑔(0)
Hence at 𝑥 = 0, the function 𝑔 is continuous.
Third case: If 𝑘 > 0,
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(𝑥) = 0. Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers greater than 0.
Hence, the function 𝑔 is continuous for all real numbers.
Therefore, 𝑔 and ℎ both are continuous function. Hence, 𝑓 is continuous.
34. Find all the points of discontinuity of 𝑓 defined by 𝑓(𝑥) = |𝑥| − |𝑥 + 1|.
Solution:
Given that the function is defined by 𝑓(𝑥) = |𝑥| − |𝑥 + 1|
Assuming that the functions are well defined for all real numbers, we can write the given function 𝑓 in the combination of 𝑔 and ℎ(𝑓 = 𝑔 − ℎ), where, 𝑔(𝑥) = |𝑥| and ℎ(𝑥) =|𝑥 + 1|. If 𝑔 and ℎ both are continuous function then 𝑓 also be continuous.
Function 𝑔(𝑥) = |𝑥|
Redefining the function 𝑔, we get,
𝑔(𝑥) = {−𝑥, If 𝑥 < 0𝑥, If 𝑥 ≥ 0
Let 𝑘 be any real number. According to question, 𝑘 < 0 or 𝑘 = 0 or 𝑘 > 0
First case: If 𝑘 < 0,
-
Class-XII-Maths Continuity and Differentiabil ity
24 Practice more on Continuity and Differentiability www.embibe.com
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(−𝑥) = 0. Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers less than 0.
Second case: If 𝑘 = 0, 𝑔(0) = 0 + 1 = 1
LHL = lim𝑥→0−
𝑔(𝑥) = lim𝑥→0−
(−𝑥) = 0 and RHL = lim𝑥→0+
𝑔(𝑥) = lim𝑥→0+
(𝑥) = 0
Here, at 𝑥 = 0, LHL = RHL = g(0)
Hence, the function 𝑔 is continuous at 𝑥 = 0.
Third case: If 𝑘 > 0,
𝑔(𝑘) = 0 and lim𝑥→𝑘
𝑔(𝑥) = lim𝑥→𝑘
(𝑥) = 0. Here, lim𝑥→𝑘
𝑔(𝑥) = 𝑔(𝑘)
Hence, the function 𝑔 is continuous for all real numbers more than 0.
Hence, the function 𝑔 is continuous for all real numbers.
Function ℎ(𝑥) = |𝑥 + 1|
Redefining the function ℎ, we get
ℎ(𝑥) = {−(𝑥 + 1), If 𝑥 < −1
𝑥 + 1, If 𝑥 ≥ −1
Let 𝑘 be any real number. According to question, 𝑘 < −1 or 𝑘 = −1 or 𝑘 > −1
First case: If 𝑘 < −1,
ℎ(𝑘) = −(𝑘 + 1) and lim𝑥→𝑘
ℎ(𝑥) = lim𝑥→𝑘
−(𝑘 + 1) = −(𝑘 + 1). Here, lim𝑥→𝑘
ℎ(𝑥) = ℎ(𝑘)
Hence, the function 𝑔 is continuous for all real numbers less than −1.
Second case: If 𝑘 = −1, ℎ(−1) = −1 + 1 = 0
LHL = lim𝑥→−1−
ℎ(𝑥) = lim𝑥→−1−
−(−1 + 1) = 0
RHL = lim𝑥→−1+
ℎ(𝑥) = lim𝑥→−1+
(𝑥 + 1) = −1 + 1 = 0
Here at 𝑥 = −1, LHL = RHL = ℎ(−1)
Hence, the function ℎ is continuous at 𝑥 = −1.
Third case: If 𝑘 > −1
ℎ(𝑘) = 𝑘 + 1 and lim𝑥→𝑘
ℎ(𝑥) = lim𝑥→𝑘
(𝑘 + 1) = 𝑘 + 1. Here, lim𝑥→𝑘
ℎ(𝑥) = ℎ(𝑘)
Hence, the function 𝑔 is continuous for all real numbers greater than −1.
Hence, the function ℎ is continuous for all real numbers.
Therefore, 𝑔 and ℎ both are continuous function. Hence, 𝑓 is continuous.
-
Class-XII-Maths Continuity and Differentiabil ity
25 Practice more on Continuity and Differentiability www.embibe.com
Exercise 𝟓. 𝟐
1. Differentiate the functions with respect to 𝑥
sin(𝑥2 + 5)
Solution:
Let 𝑦 = sin(𝑥2 + 5)
Therefore,
𝑑𝑦
𝑑𝑥= cos(𝑥2 + 5).
𝑑
𝑑𝑥(𝑥2 + 5)
= cos(𝑥2 + 5). 2𝑥
Hence, 𝑑(sin(𝑥2+5))
𝑑𝑥 = cos(𝑥2 + 5). 2𝑥
2. Differentiate the functions with respect to 𝑥
cos(sin 𝑥)
Solution:
Let 𝑦 = cos(sin 𝑥)
Therefore,
𝑑𝑦
𝑑𝑥= −sin(sin𝑥).
𝑑
𝑑𝑥(sin 𝑥)
= −sin(sin𝑥) . cos 𝑥
Hence, 𝑑((cos(sin𝑥)))
𝑑𝑥 = − sin(sin𝑥) . cos 𝑥.
3. Differentiate the functions with respect to 𝑥
sin(𝑎𝑥 + 𝑏)
Solution:
Let 𝑦 = sin(𝑎𝑥 + 𝑏)
Therefore,
𝑑𝑦
𝑑𝑥= cos(𝑎𝑥 + 𝑏).
𝑑
𝑑𝑥(𝑎𝑥 + 𝑏)
-
Class-XII-Maths Continuity and Differentiabil ity
26 Practice more on Continuity and Differentiability www.embibe.com
= cos(𝑎𝑥 + 𝑏). 𝑎
Hence, 𝑑(sin(𝑎𝑥+𝑏))
𝑑𝑥= cos(𝑎𝑥 + 𝑏). 𝑎
4. Differentiate the functions with respect to 𝑥
sec(tan(√𝑥))
Solution:
Let 𝑦 = sec(tan(√𝑥))
Therefore,
𝑑𝑦
𝑑𝑥= sec(tan√𝑥) tan(tan√𝑥).
𝑑
𝑑𝑥(tan√𝑥)
= sec(tan√𝑥) tan(tan√𝑥). sec2√𝑥𝑑
𝑑𝑥(√𝑥)
= sec(tan√𝑥) tan(tan√𝑥) . sec2√𝑥 (1
2√𝑥)
Hence, 𝑑(sec(tan(√𝑥)))
𝑑𝑥= sec(tan√𝑥) tan(tan√𝑥) . sec2√𝑥 (
1
2√𝑥)
5. Differentiate the functions with respect to 𝑥
sin(𝑎𝑥+𝑏)
cos(𝑐𝑥+𝑑)
Solution:
Let 𝑦 =sin(𝑎𝑥+𝑏)
cos(𝑐𝑥+𝑑)
Therefore,
𝑑𝑦
𝑑𝑥=cos(𝑐𝑥 + 𝑑).
𝑑𝑑𝑥sin(𝑎𝑥 + 𝑏) − sin(𝑎𝑥 + 𝑏).
𝑑𝑑𝑥cos(𝑐𝑥 + 𝑑)
[cos(𝑐𝑥 + 𝑑)]2
=cos(𝑐𝑥 + 𝑑). sin(𝑎𝑥 + 𝑏).
𝑑𝑑𝑥(𝑎𝑥 + 𝑏) − sin(𝑎𝑥 + 𝑏). [− sin(𝑐𝑥 + 𝑑).
𝑑𝑑𝑥(𝑐𝑥 + 𝑑)]
cos2(𝑐𝑥 + 𝑑)
=cos(𝑐𝑥+𝑑).sin(𝑎𝑥+𝑏).𝑎+sin(𝑎𝑥+𝑏).sin(𝑐𝑥+𝑑)𝑐
cos2(𝑐𝑥+𝑑)
Hence, 𝑑(
sin(𝑎𝑥+𝑏)
cos(𝑐𝑥+𝑑))
𝑑𝑥=
cos(𝑐𝑥+𝑑).sin(𝑎𝑥+𝑏).𝑎+sin(𝑎𝑥+𝑏).sin(𝑐𝑥+𝑑)𝑐
cos2(𝑐𝑥+𝑑)
-
Class-XII-Maths Continuity and Differentiabil ity
27 Practice more on Continuity and Differentiability www.embibe.com
6. Differentiate the functions with respect to 𝑥
cos 𝑥3. sin2(𝑥5)
Solution:
Let 𝑦 = cos 𝑥3. sin2(𝑥5)
Therefore,
𝑑𝑦
𝑑𝑥= cos 𝑥3.
𝑑
𝑑𝑥sin2(𝑥5) + sin2(𝑥5).
𝑑
𝑑𝑥cos𝑥3
= cos 𝑥3. 2 sin 𝑥5 cos𝑥5.𝑑
𝑑𝑥𝑥5 + sin2(𝑥5)[− sin 𝑥3].
𝑑
𝑑𝑥𝑥3
= cos 𝑥3. 2 sin 𝑥5 cos𝑥5. 5𝑥4 − sin2(𝑥5) sin 𝑥3. 3𝑥2
Hence, 𝑑(cos𝑥3.sin2(𝑥5))
𝑑𝑥= cos 𝑥3. 2 sin 𝑥5 cos 𝑥5. 5𝑥4 − sin2(𝑥5) sin 𝑥3. 3𝑥2
7. Differentiate the functions with respect to 𝑥
2√cot(𝑥2)
Solution:
Let 𝑦 = 2√cot(𝑥2)
Therefore,
𝑑𝑦
𝑑𝑥= 2.
1
2√cot(𝑥2).𝑑
𝑑𝑥[cot(𝑥2)]
=1
√cot(𝑥2). [− cossec 𝑥2].
𝑑
𝑑𝑥𝑥2
=1
√cot(𝑥2). [− cossec 𝑥2]. 2𝑥
Hence, 𝑑(2√cot(𝑥2))
𝑑𝑥=
1
√cot(𝑥2). [− cossec 𝑥2]. 2𝑥
8. Differentiate the functions with respect to 𝑥
cos(√𝑥)
-
Class-XII-Maths Continuity and Differentiabil ity
28 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Let 𝑦 = cos(√𝑥)
Therefore,
𝑑𝑦
𝑑𝑥= −sin(√𝑥).
𝑑
𝑑𝑥√𝑥
= −sin(√𝑥).1
2√𝑥
Hence, 𝑑(cos(√𝑥))
𝑑𝑥= − sin(√𝑥).
1
2√𝑥
9. Prove that the function 𝑓 given by 𝑓(𝑥) = |𝑥 − 1|, 𝑥 ∈ 𝑅, is not differentiable at 𝑥 = 1.
Solution:
At 𝑥 = 1,
LHD = limℎ→0
𝑓(1 − ℎ) − 𝑓(1)
−ℎ= lim
ℎ→0
|1 − ℎ − 1| − |1 − 1|
−ℎ= limℎ→0
ℎ
−ℎ= −1
RHD = limℎ→0
𝑓(1+ℎ)−𝑓(1)
ℎ= lim
ℎ→0
|1+ℎ−1|−|1−1|
ℎ= lim
ℎ→0
ℎ
ℎ= 1
Here, LHD ≠ RHD, therefore,
the function 𝑓(𝑥) = |𝑥 − 1|, 𝑥 ∈ 𝑅, is not differentiable at 𝑥 = 1.
10. Prove that the greatest integer function defined by 𝑓(𝑥) = [𝑥], 0 < 𝑥 < 3, is not differentiable at 𝑥 = 1 and 𝑥 = 2.
Solution:
At 𝑥 = 1,
LHD = limℎ→0
𝑓(1 − ℎ) − 𝑓(1)
−ℎ= lim
ℎ→0
[1 − ℎ] − |1|
−ℎ= limℎ→0
0 − 1
−ℎ= ∞
RHD = limℎ→0
𝑓(1 + ℎ) − 𝑓(1)
ℎ= lim
ℎ→0
[1 + ℎ] − [1]
ℎ= limℎ→0
1 − 1
ℎ= 0
Here, LHD ≠ RHD, therefore,
The function 𝑓(𝑥) = [𝑥], 0 < 𝑥 < 3, is not differentiable at 𝑥 = 1.
At 𝑥 = 2,
-
Class-XII-Maths Continuity and Differentiabil ity
29 Practice more on Continuity and Differentiability www.embibe.com
LHD = limℎ→0
𝑓(1 − ℎ) − 𝑓(1)
−ℎ= lim
ℎ→0
[2 − ℎ] − [2]
−ℎ= lim
ℎ→0
1 − 2
−ℎ= ∞
RHD = limℎ→0
𝑓(1 + ℎ) − 𝑓(1)
ℎ= lim
ℎ→0
[2 + ℎ] − [2]
ℎ= lim
ℎ→0
2 − 2
ℎ= 0
Here, LHD ≠ RHD, therefore,
The function 𝑓(𝑥) = [𝑥], 0 < 𝑥 < 3, is not differentiable at 𝑥 = 2.
Exercise 𝟓. 𝟑
Find 𝑑𝑦
𝑑𝑥 in the following:
1. 2𝑥 + 3𝑦 = sin 𝑥
Solution:
Given equation is 2𝑥 + 3𝑦 = sin𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥(2𝑥) +
𝑑
𝑑𝑥(3𝑦) =
𝑑
𝑑𝑥sin𝑥
⇒ 2+ 3𝑑𝑦
𝑑𝑥= cos𝑥
⇒𝑑𝑦
𝑑𝑥=
cos𝑥−2
3
2. 2𝑥 + 3𝑦 = sin 𝑦
Solution:
Given equation is 2𝑥 + 3𝑦 = sin𝑦
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥(2𝑥) +
𝑑
𝑑𝑥(3𝑦) =
𝑑
𝑑𝑥sin𝑦 ⇒ 2 + 3
𝑑𝑦
𝑑𝑥= cos 𝑦
𝑑𝑦
𝑑𝑥
⇒𝑑𝑦
𝑑𝑥(cos𝑦 − 3) = 2
⇒𝑑𝑦
𝑑𝑥=
2
cos𝑦−3
-
Class-XII-Maths Continuity and Differentiabil ity
30 Practice more on Continuity and Differentiability www.embibe.com
3. 𝑎𝑥 + 𝑏𝑦2 = cos 𝑦
Solution:
Given equation is 𝑎𝑥 + 𝑏𝑦2 = cos 𝑦
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥(𝑎𝑥) +
𝑑
𝑑𝑥(𝑏𝑦2) =
𝑑
𝑑𝑥cos 𝑦 ⇒ 𝑎 + 2𝑏𝑦
𝑑𝑦
𝑑𝑥= −sin𝑦
𝑑𝑦
𝑑𝑥
⇒𝑑𝑦
𝑑𝑥(2𝑏𝑦 + sin𝑦) = −𝑎 ⇒
𝑑𝑦
𝑑𝑥= −
𝑎
2𝑏𝑦+sin𝑦
4. 𝑥𝑦 + 𝑦2 = tan 𝑥 + 𝑦
Solution:
Given equation is 𝑥𝑦 + 𝑦2 = tan𝑥 + 𝑦
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥(𝑥𝑦) +
𝑑
𝑑𝑥(𝑦2) =
𝑑
𝑑𝑥tan 𝑥 +
𝑑𝑦
𝑑𝑥
⇒ 𝑥𝑑𝑦
𝑑𝑥+ 𝑦 + 2𝑦
𝑑𝑦
𝑑𝑥= sec2 𝑥 +
𝑑𝑦
𝑑𝑥
⇒𝑑𝑦
𝑑𝑥(𝑥 + 2𝑦 − 1) = sec2 𝑥 − 𝑦
⇒𝑑𝑦
𝑑𝑥=
sec2 𝑥−𝑦
𝑥+2𝑦−1
5. 𝑥2 + 𝑥𝑦 + 𝑦2 = 100
Solution:
Given equation is 𝑥2 + 𝑥𝑦 + 𝑦2 = 100
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥𝑥2 +
𝑑
𝑑𝑥(𝑥𝑦) +
𝑑
𝑑𝑥𝑦2 =
𝑑
𝑑𝑥(100)
⇒ 2𝑥 + 𝑥𝑑𝑦
𝑑𝑥+ 𝑦 + 2𝑦
𝑑𝑦
𝑑𝑥= 0
⇒𝑑𝑦
𝑑𝑥(𝑥 + 2𝑦) = 2𝑥 + 𝑦 ⇒
𝑑𝑦
𝑑𝑥=
2𝑥+𝑦
𝑥+2𝑦
-
Class-XII-Maths Continuity and Differentiabil ity
31 Practice more on Continuity and Differentiability www.embibe.com
6. 𝑥3 + 𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 = 81
Solution:
Given equation is 𝑥3 + 𝑥2𝑦 + 𝑥𝑦2 + 𝑦3 = 81
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥𝑥3 +
𝑑
𝑑𝑥(𝑥2𝑦) +
𝑑
𝑑𝑥(𝑥𝑦2) +
𝑑
𝑑𝑥𝑦3 =
𝑑
𝑑𝑥81
⇒ 3𝑥2 + 𝑥2𝑑𝑦
𝑑𝑥+ 𝑦. 2𝑥 + 𝑥. 2𝑦
𝑑𝑦
𝑑𝑥+ 𝑦2. 1 + 3𝑦2
𝑑𝑦
𝑑𝑥= 0
⇒𝑑𝑦
𝑑𝑥(𝑥2 + 2𝑥𝑦 + 3𝑦2) = −(3𝑥2 + 2𝑥𝑦 + 𝑦2) ⇒
𝑑𝑦
𝑑𝑥= −
3𝑥2+2𝑥𝑦+𝑦2
𝑥2+2𝑥𝑦+3𝑦2
7. sin2 𝑦 + cos𝑥𝑦 = 𝑘
Solution:
Given equation is sin2 𝑦 + cos 𝑥𝑦 = 𝑘
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥sin2 𝑦 +
𝑑
𝑑𝑥cos 𝑥𝑦 =
𝑑
𝑑𝑥𝑘
⇒ 2sin𝑦 cos 𝑦𝑑𝑦
𝑑𝑥− sin𝑥𝑦 (𝑥
𝑑𝑦
𝑑𝑥+ 𝑦) = 0
⇒ sin2𝑦𝑑𝑦
𝑑𝑥− 𝑥 sin𝑥𝑦
𝑑𝑦
𝑑𝑥− 𝑦 sin 𝑥𝑦 = 0
⇒ (sin2𝑦 − 𝑥 sin𝑥𝑦)𝑑𝑦
𝑑𝑥= 𝑦 sin𝑥𝑦
⇒𝑑𝑦
𝑑𝑥=
𝑦 sin𝑥𝑦
sin2𝑦−𝑥 sin𝑥𝑦
8. sin2 𝑥 + cos2 𝑦 = 1
Solution:
Given equation is sin2 𝑥 + cos2 𝑦 = 1
Differentiating both sides with respect to 𝑥, we get
𝑑
𝑑𝑥sin2 𝑥 +
𝑑
𝑑𝑥cos2 𝑦 =
𝑑
𝑑𝑥1
⇒ 2sin𝑥 cos 𝑥 + 2 cos𝑦 (− sin𝑦)𝑑𝑦
𝑑𝑥= 0
-
Class-XII-Maths Continuity and Differentiabil ity
32 Practice more on Continuity and Differentiability www.embibe.com
⇒ sin2𝑥 − sin2𝑦𝑑𝑦
𝑑𝑥= 0 ⇒
𝑑𝑦
𝑑𝑥=
sin2𝑥
sin2𝑦
9. 𝑦 = sin−1 (2𝑥
1+𝑥2)
Solution:
Given equation is 𝑦 = sin−1 (2𝑥
1+𝑥2)
Let 𝑥 = tan𝜃
Therefore, 𝑦 = sin−1 (2 tan𝜃
1+tan2 𝜃) = sin−1(sin 2𝜃) = 2𝜃 = 2 tan−1 𝑥
⇒ 𝑦 = 2 tan−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
2
1+𝑥2
10. 𝑦 = tan−1 (3𝑥−𝑥3
1−3𝑥2) , −
1
√3< 𝑥 <
1
√3
Solution:
Given equation is 𝑦 = tan−1 (3𝑥−𝑥3
1−3𝑥2)
Let 𝑥 = tan𝜃
Therefore, 𝑦 = tan−1 (3 tan𝜃−tan3 𝜃
1−3 tan2 𝜃)
= tan−1(tan 3𝜃) = 3𝜃 = 3 tan−1 𝑥 ⇒ 𝑦 = 3 tan−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
3
1+𝑥2
11. 𝑦 = cos−1 (1−𝑥2
1+𝑥2) , 0 < 𝑥 < 1
-
Class-XII-Maths Continuity and Differentiabil ity
33 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given equation is 𝑦 = cos−1 (1−𝑥2
1+𝑥2)
Let 𝑥 = tan𝜃
Therefore, 𝑦 = cos−1 (1−tan2 𝜃
1+tan2 𝜃)
= cos−1(cos 2𝜃) = 2𝜃 = 2 tan−1 𝑥
⇒ 𝑦 = 2 tan−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
2
1+𝑥2
12. 𝑦 = sin−1 (1−𝑥2
1+𝑥2) , 0 < 𝑥 < 1
Solution:
Given equation is 𝑦 = sin−1 (1−𝑥2
1+𝑥2)
Let 𝑥 = tan𝜃
Therefore,
𝑦 = sin−1 (1−tan2 𝜃
1+tan2 𝜃)
= sin−1(cos 2𝜃) = sin−1 {sin (𝜋
2− 2𝜃)} =
𝜋
2− 2𝜃 =
𝜋
2− 2 tan−1 𝑥
⇒ 𝑦 =𝜋
2− 2 tan−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥= 0 −
2
1+𝑥2= −
2
1+𝑥2
13. 𝑦 = cos−1 (2𝑥
1+𝑥2) , −1 < 𝑥 < 1
Solution:
Given equation is 𝑦 = cos−1 (2𝑥
1+𝑥2)
Let 𝑥 = tan𝜃
Therefore, 𝑦 = cos−1 (2 tan𝜃
1+tan2 𝜃)
-
Class-XII-Maths Continuity and Differentiabil ity
34 Practice more on Continuity and Differentiability www.embibe.com
= cos−1(sin 2𝜃) = cos−1 {cos (𝜋
2− 2𝜃)} =
𝜋
2− 2𝜃 =
𝜋
2− 2 tan−1 𝑥
⇒ 𝑦 =𝜋
2− 2 tan−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥= 0 −
2
1+𝑥2= −
2
1+𝑥2
14. 𝑦 = sin−1(2𝑥√1 − 𝑥2),−1
√2< 𝑥 <
1
√2
Solution:
Given equation is 𝑦 = sin−1(2𝑥√1 − 𝑥2)
Let 𝑥 = sin 𝜃
Therefore, 𝑦 = sin−1(2 sin𝜃√1 − sin2 𝜃)
= sin−1(2 sin𝜃 cos𝜃) = sin−1(sin2𝜃) = 2𝜃 = 2 sin−1 𝑥
⇒ 𝑦 = 2 sin−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
2
√1−𝑥2
15. 𝑦 = sec−1 (1
2𝑥2−1) , 0 < 𝑥 <
1
√2
Solution:
Given equation is 𝑦 = sec−1 (1
2𝑥2−1)
Let 𝑥 = cos𝜃
Therefore, 𝑦 = sec−1 (1
2 cos2 𝜃−1) = sec−1 (
1
cos2𝜃)
sec−1(sec 2𝜃) = 2𝜃 = 2 cos−1 𝑥
⇒ 𝑦 = 2 cos−1 𝑥
Differentiating both sides with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥= −
2
√1−𝑥2
-
Class-XII-Maths Continuity and Differentiabil ity
35 Practice more on Continuity and Differentiability www.embibe.com
Exercise 𝟓. 𝟒
1. Differentiate the following w.r.t. 𝑥:
𝑒𝑥
sin𝑥
Solution:
Given expression is 𝑒𝑥
sin𝑥
Let 𝑦 =𝑒𝑥
sin𝑥 therefore,
𝑑𝑦
𝑑𝑥=
𝑒𝑥.𝑑
𝑑𝑥sin𝑥−sin𝑥
𝑑
𝑑𝑥𝑒𝑥
sin2 𝑥=
𝑒𝑥.cos𝑥−sin𝑥.𝑒𝑥
sin2 𝑥=
𝑒𝑥(cos𝑥−sin𝑥)
sin2 𝑥
2. 𝑒sin−1 𝑥
Solution:
Given expression is 𝑒sin−1 𝑥
Let 𝑦 = 𝑒sin−1 𝑥 , therefore,
𝑑𝑦
𝑑𝑥= 𝑒sin
−1 𝑥 .𝑑
𝑑𝑥sin−1 𝑥 = 𝑒sin
−1 𝑥 .1
√1−𝑥2=
𝑒sin−1 𝑥
√1−𝑥2.
3. 𝑒𝑥3
Solution:
Given expression is 𝑒𝑥3
Let 𝑦 = 𝑒𝑥3, therefore,
𝑑𝑦
𝑑𝑥= 𝑒𝑥
3.𝑑
𝑑𝑥𝑥3 = 𝑒𝑥
3. 3𝑥2 = 3𝑥2𝑒𝑥
3
-
Class-XII-Maths Continuity and Differentiabil ity
36 Practice more on Continuity and Differentiability www.embibe.com
4. sin(tan−1 𝑒−𝑥)
Solution:
Given expression is sin(tan−1 𝑒−𝑥)
Let 𝑦 = sin(tan−1 𝑒−𝑥), therefore,
𝑑𝑦
𝑑𝑥= cos(tan−1 𝑒−𝑥) .
𝑑
𝑑𝑥tan−1 𝑒−𝑥 = cos(tan−1 𝑒−𝑥) .
1
1+(𝑒−𝑥)2.𝑑
𝑑𝑥𝑒−𝑥
= cos(tan−1 𝑒−𝑥) .1
1+𝑒−2𝑥. (−𝑒−𝑥) = −
𝑒−𝑥 cos(tan−1 𝑒−𝑥)
1+𝑒−2𝑥.
5. log(cos 𝑒𝑥)
Solution:
Given expression is log(cos 𝑒𝑥)
Let 𝑦 = log(cos 𝑒𝑥),
Therefore,
𝑑𝑦
𝑑𝑥=
1
cos𝑒𝑥.𝑑
𝑑𝑥cos 𝑒𝑥 =
1
cos𝑒𝑥(− sin 𝑒𝑥)
𝑑
𝑑𝑥𝑒𝑥 = − tan 𝑒𝑥. 𝑒𝑥
6. 𝑒𝑥 + 𝑒𝑥2+⋯+ 𝑒𝑥
5
Solution:
Given expression is 𝑒𝑥 + 𝑒𝑥2+⋯+ 𝑒𝑥
5
Let 𝑦 = 𝑒𝑥 + 𝑒𝑥2+ 𝑒𝑥
3+ 𝑒𝑥
4+ 𝑒𝑥
5, therefore,
𝑑𝑦
𝑑𝑥= 𝑒𝑥 + 𝑒𝑥
2 𝑑
𝑑𝑥𝑥2 + 𝑒𝑥
3 𝑑
𝑑𝑥𝑥3 + 𝑒𝑥
4 𝑑
𝑑𝑥𝑥4 + 𝑒𝑥
5 𝑑
𝑑𝑥𝑥5
= 𝑒𝑥 + 𝑒𝑥2. 2𝑥 + 𝑒𝑥
3. 3𝑥2 + 𝑒𝑥
4. 4𝑥3 + 𝑒𝑥
5. 5𝑥4
= 𝑒𝑥 + 2𝑥𝑒𝑥2+ 3𝑥2𝑒𝑥
3+ 4𝑥3𝑒𝑥
4+ 5𝑥4𝑒𝑥
5
7. √𝑒√𝑥, 𝑥 > 0
-
Class-XII-Maths Continuity and Differentiabil ity
37 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given expression is √𝑒√𝑥, 𝑥 > 0
Let 𝑦 = √𝑒√𝑥
Therefore,
𝑑𝑦
𝑑𝑥=
1
2√𝑒√𝑥
𝑑
𝑑𝑥𝑒√𝑥 =
1
2√𝑒√𝑥. 𝑒√𝑥.
𝑑
𝑑𝑥√𝑥 =
1
2√𝑒√𝑥. 𝑒√𝑥.
1
2√𝑥=
√𝑒√𝑥
4√𝑥
8. log(log 𝑥), 𝑥 > 1
Solution:
Given expression is log(log 𝑥), 𝑥 > 1
Let 𝑦 =𝑒𝑥
sin𝑥
Therefore,
𝑑𝑦
𝑑𝑥=
1
log𝑥.𝑑
𝑑𝑥log 𝑥 =
1
log𝑥.1
𝑥=
1
𝑥 log𝑥
9. cos𝑥
log𝑥, 𝑥 > 0
Solution:
Given expression is cos𝑥
log 𝑥, 𝑥 > 0
Let 𝑦 =cos𝑥
log𝑥
Therefore, 𝑑𝑦
𝑑𝑥=
log 𝑥𝑑
𝑑𝑥cos𝑥−cos𝑥
𝑑
𝑑𝑥log𝑥
(log𝑥)2=
log𝑥.(−sin𝑥)−cos𝑥.1
𝑥
(log𝑥)2=
−(𝑥 sin𝑥 log 𝑥+cos𝑥)
𝑥(log𝑥)2
10. cos(log 𝑥 + 𝑒𝑥)
Solution:
Given expression is cos(log 𝑥 + 𝑒𝑥)
-
Class-XII-Maths Continuity and Differentiabil ity
38 Practice more on Continuity and Differentiability www.embibe.com
Let 𝑦 = cos(log 𝑥 + 𝑒𝑥)
Therefore,
𝑑𝑦
𝑑𝑥= −sin(log 𝑥 + 𝑒𝑥).
𝑑
𝑑𝑥(log 𝑥 + 𝑒𝑥) = − sin(log 𝑥 + 𝑒𝑥). (
1
𝑥+ 𝑒𝑥)
Exercise 𝟓. 𝟓
1. Differentiate the functions given
cos 𝑥. cos 2𝑥. cos 3𝑥
Solution:
Given function is cos 𝑥. cos 2𝑥. cos 3𝑥
Let 𝑦 = cos 𝑥. cos 2𝑥. cos 3𝑥, taking log on both the sides
log 𝑦 = log cos 𝑥 + log cos 2𝑥 + log cos3𝑥
Therefore,
1
𝑦
𝑑𝑦
𝑑𝑥=
1
cos𝑥.𝑑
𝑑𝑥cos 𝑥 +
1
cos2𝑥.𝑑
𝑑𝑥cos 2𝑥 +
1
cos3𝑥.𝑑
𝑑𝑥cos3𝑥
⇒𝑑𝑦
𝑑𝑥= 𝑦 [
1
cos𝑥. (− sin𝑥) +
1
cos2𝑥. (− sin2𝑥). 2 +
1
cos3𝑥. (− sin 3𝑥). 3]
⇒𝑑𝑦
𝑑𝑥= cos𝑥. cos 2𝑥. cos 3𝑥 [− tan 𝑥 − 2 tan 2𝑥 − 3 tan3𝑥]
2. Differentiate the functions given
√(𝑥−1)(𝑥−2)
(𝑥−3)(𝑥−4)(𝑥−5)
Solution:
Given function is √(𝑥−1)(𝑥−2)
(𝑥−3)(𝑥−4)(𝑥−5)
Let 𝑦 = √(𝑥−1)(𝑥−2)
(𝑥−3)(𝑥−4)(𝑥−5), taking log on both the sides
log 𝑦 =1
2[log(𝑥 − 1) + log(𝑥 − 2) − log(𝑥 − 3) − log(𝑥 − 4) − log(𝑥 − 5)]
-
Class-XII-Maths Continuity and Differentiabil ity
39 Practice more on Continuity and Differentiability www.embibe.com
Therefore,
1
𝑦
𝑑𝑦
𝑑𝑥=
1
2[
1
(𝑥−1)+
1
(𝑥−2)−
1
(𝑥−3)−
1
(𝑥−4)−
1
(𝑥−5)]
⇒𝑑𝑦
𝑑𝑥=
1
2√
(𝑥−1)(𝑥−2)
(𝑥−3)(𝑥−4)(𝑥−5)[
1
(𝑥−1)+
1
(𝑥−2)−
1
(𝑥−3)−
1
(𝑥−4)−
1
(𝑥−5)]
3. Differentiate the functions given
(log 𝑥)cos𝑥
Solution:
Given function is (log 𝑥)cos𝑥
Let 𝑦 = (log 𝑥)cos𝑥 , taking log on both the sides
log 𝑦 = log(log 𝑥)cos𝑥 = cos 𝑥. log log 𝑥
Therefore,
1
𝑦
𝑑𝑦
𝑑𝑥= cos 𝑥.
𝑑
𝑑𝑥log log 𝑥 + log log 𝑥.
𝑑
𝑑𝑥cos 𝑥
⇒𝑑𝑦
𝑑𝑥= 𝑦 [cos 𝑥.
1
log𝑥.1
𝑥+ log log 𝑥. (− sin𝑥)]
⇒𝑑𝑦
𝑑𝑥= (log 𝑥)cos𝑥 [
cos𝑥−sin𝑥 log log𝑥
𝑥 log 𝑥]
4. Differentiate the functions given
𝑥𝑥 − 2sin𝑥
Solution:
Given function is 𝑥𝑥 − 2sin𝑥
Let 𝑢 = 𝑥𝑥 and 𝑣 = 2sin𝑥 therefore, 𝑦 = 𝑢 − 𝑣
Differentiating with respect to 𝑥 on both sides
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥−𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = 𝑥𝑥 , taking log on both the sides
-
Class-XII-Maths Continuity and Differentiabil ity
40 Practice more on Continuity and Differentiability www.embibe.com
log 𝑢 = 𝑥 log 𝑥, therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥.
𝑑
𝑑𝑥log 𝑥 + log 𝑥.
𝑑
𝑑𝑥𝑥 = 𝑥.
1
𝑥+ log 𝑥. 1 = 1 + log 𝑥
𝑑𝑢
𝑑𝑥= 𝑢[1 + log 𝑥] = 𝑥𝑥[1 + log 𝑥] …(ii)
and 𝑣 = 2sin𝑥 , taking log on both the sides
log 𝑣 = sin𝑥 log 2, therefore,
1
𝑣
𝑑𝑣
𝑑𝑥= log2 ⋅
𝑑
𝑑𝑥sin 𝑥 = log 2 ⋅ cos 𝑥
𝑑𝑣
𝑑𝑥= 𝑣[cos 𝑥 log 2] = 2sin𝑥[cos 𝑥 log 2] …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= 𝑥𝑥[1 + log 𝑥] − 2sin𝑥[cos 𝑥 log 2]
5. Differentiate the functions given
(𝑥 + 3)2. (𝑥 + 4)3. (𝑥 + 5)4
Solution:
Given function is (𝑥 + 3)2. (𝑥 + 4)3. (𝑥 + 5)4
Let 𝑦 = (𝑥 + 3)2. (𝑥 + 4)3. (𝑥 + 5)4, taking log on both the sides
log 𝑦 = 2 log(𝑥 + 3) + 3 log(𝑥 + 4) + 4 log(𝑥 + 5)
Therefore,
1
𝑦
𝑑𝑦
𝑑𝑥= 2.
1
(𝑥+3)+ 3.
1
(𝑥+4)+ 4.
1
(𝑥+5)
⇒𝑑𝑦
𝑑𝑥= 𝑦 [
2(𝑥 + 4)(𝑥 + 5) + 3(𝑥 + 3)(𝑥 + 5) + 4(𝑥 + 3)(𝑥 + 4)
(𝑥 + 3)(𝑥 + 4)(𝑥 + 5)]
⇒𝑑𝑦
𝑑𝑥= 𝑦 [
2(𝑥2 + 9𝑥 + 20) + 3(𝑥2 + 8𝑥 + 15) + 4(𝑥2 + 7𝑥 + 12)
(𝑥 + 3)(𝑥 + 4)(𝑥 + 5)]
⇒𝑑𝑦
𝑑𝑥= (𝑥 + 3)2 ⋅ (𝑥 + 4)3 ⋅ (𝑥 + 5)4 [
9𝑥2+70𝑥+133
(𝑥+3)(𝑥+4)(𝑥+5)]
⇒𝑑𝑦
𝑑𝑥= (𝑥 + 3) ⋅ (𝑥 + 4)2 ⋅ (𝑥 + 5)3(9𝑥2 + 70𝑥 + 133)
6. Differentiate the functions given
(𝑥 +1
𝑥)𝑥+ 𝑥
(1+1
𝑥)
-
Class-XII-Maths Continuity and Differentiabil ity
41 Practice more on Continuity and Differentiability www.embibe.com
Solution:
Given function is (𝑥 +1
𝑥)𝑥+ 𝑥
(1+1
𝑥)
Let 𝑢 = (𝑥 +1
𝑥)𝑥
and 𝑣 = 𝑥(1+
1
𝑥), therefore, 𝑦 = 𝑢 + 𝑣
Differentiating with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = (𝑥 +1
𝑥)𝑥, taking log on both the sides
log 𝑢 = 𝑥 log (𝑥 +1
𝑥), therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥 ⋅
𝑑
𝑑𝑥log (𝑥 +
1
𝑥) + log (𝑥 +
1
𝑥) ⋅
𝑑
𝑑𝑥𝑥
= 𝑥 ⋅1
(𝑥+1
𝑥)⋅ (1 −
1
𝑥2) + log (𝑥 +
1
𝑥) ⋅ 1 =
𝑥2
𝑥2+1⋅𝑥2−1
𝑥2+ log (𝑥 +
1
𝑥)
𝑑𝑢
𝑑𝑥= (𝑥 +
1
𝑥)𝑥[𝑥2−1
𝑥2+1+ log (𝑥 +
1
𝑥)] …(ii)
and 𝑣 = 𝑥(1+
1
𝑥), taking log on both the sides
log 𝑣 = (1 +1
𝑥) log 𝑥, therefore,
1
𝑣
𝑑𝑣
𝑑𝑥= (1 +
1
𝑥) ·
𝑑
𝑑𝑥log 𝑥 + log 𝑥 ·
𝑑
𝑑𝑥(1 +
1
𝑥) = (1 +
1
𝑥) ·
1
𝑥+ log 𝑥 · (−
1
𝑥2)
𝑑𝑣
𝑑𝑥= 𝑣 [(
𝑥2+1
𝑥) ·
1
𝑥−log𝑥
𝑥2] = 𝑥
(1+1
𝑥)[𝑥2+1−log𝑥
𝑥2] …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= (𝑥 +
1
𝑥)𝑥[𝑥2−1
𝑥2+1+ log (𝑥 +
1
𝑥)] + 𝑥
(1+1
𝑥)[𝑥2+1−log𝑥
𝑥2]
7. Differentiate the functions given
(log 𝑥)𝑥 + 𝑥log𝑥
Solution:
Given function is (log 𝑥)𝑥 + 𝑥log 𝑥
Let 𝑢 = (log 𝑥)𝑥 and 𝑣 = 𝑥log𝑥 , therefore, 𝑦 = 𝑢 + 𝑣
Differentiating with respect to 𝑥, we get
-
Class-XII-Maths Continuity and Differentiabil ity
42 Practice more on Continuity and Differentiability www.embibe.com
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = (log 𝑥)𝑥, taking log on both the sides
log 𝑢 = 𝑥 log log 𝑥, therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥.
𝑑
𝑑𝑥log log 𝑥 + log log 𝑥.
𝑑
𝑑𝑥𝑥
= 𝑥.1
log 𝑥.1
𝑥+ log log 𝑥. 1 =
1
log 𝑥+ log log 𝑥
𝑑𝑢
𝑑𝑥= (log 𝑥)𝑥 [
1 + log 𝑥. log log 𝑥
log𝑥]
= (log 𝑥)𝑥−1(1 + log 𝑥. log log 𝑥) …(ii)
and, 𝑣 = 𝑥log𝑥 , taking log on both the sides
log 𝑣 = log 𝑥 log 𝑥, therefore,
1
𝑣
𝑑𝑣
𝑑𝑥= log 𝑥.
𝑑
𝑑𝑥log 𝑥 + 𝑙𝑜𝑔𝑥 .
𝑑
𝑑𝑥log 𝑥
= log 𝑥.1
𝑥+ log 𝑥.
1
𝑥
𝑑𝑣
𝑑𝑥= 𝑣 [
2 log 𝑥
𝑥] = 𝑥log𝑥 [
2 log 𝑥
𝑥] = 𝑥log𝑥−1(2 log 𝑥) …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= (log 𝑥)𝑥−1(1 + log 𝑥. log log 𝑥) + 𝑥log𝑥−1(2 log 𝑥)
8. Differentiate the functions given
(sin 𝑥)𝑥 + sin−1√𝑥
Solution:
Given function is (sin 𝑥)𝑥 + sin−1 √𝑥
Let 𝑢 = (sin𝑥)𝑥 and 𝑣 = sin−1 √𝑥, therefore, 𝑦 = 𝑢 + 𝑣
Differentiating with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = (sin 𝑥)𝑥, taking log on both the sides
log 𝑢 = 𝑥 log sin 𝑥, therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥.
𝑑
𝑑𝑥log sin 𝑥 + log sin𝑥.
𝑑
𝑑𝑥𝑥
= 𝑥.1
sin 𝑥. cos 𝑥 + log sin 𝑥. 1 = 𝑥 cot 𝑥 + log sin 𝑥
-
Class-XII-Maths Continuity and Differentiabil ity
43 Practice more on Continuity and Differentiability www.embibe.com
𝑑𝑢
𝑑𝑥= (sin 𝑥)𝑥(𝑥 cot 𝑥 + log sin𝑥) …(ii)
and, 𝑣 = sin−1 √𝑥, therefore,
1
𝑣
𝑑𝑣
𝑑𝑥= log 𝑥.
𝑑
𝑑𝑥log 𝑥 + log 𝑥.
𝑑
𝑑𝑥log 𝑥
= log 𝑥.1
𝑥+ log 𝑥.
1
𝑥
𝑑𝑣
𝑑𝑥=
1
√1−𝑥.𝑑
𝑑𝑥√𝑥 =
1
√1−𝑥.1
2√𝑥=
1
2√𝑥−𝑥2 …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= (sin𝑥)𝑥(𝑥 cot 𝑥 + log sin𝑥) +
1
2√𝑥−𝑥2
9. Differentiate the functions given
𝑥sin𝑥 + (sin 𝑥)cos𝑥
Solution:
Given function is 𝑥sin𝑥 + (sin𝑥)cos𝑥
Let 𝑢 = 𝑥sin𝑥 and 𝑣 = (sin 𝑥)cos𝑥 therefore, 𝑦 = 𝑢 + 𝑣
Differentiating with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = 𝑥sin𝑥 , taking log on both the sides
log 𝑢 = sin 𝑥 log 𝑥, therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= sin𝑥.
𝑑
𝑑𝑥log 𝑥 + log 𝑥.
𝑑
𝑑𝑥sin 𝑥 = sin𝑥.
1
𝑥+ log 𝑥. cos 𝑥 =
sin𝑥
𝑥+ log 𝑥 cos 𝑥
𝑑𝑢
𝑑𝑥= 𝑥sin𝑥 [
sin𝑥
𝑥+ log 𝑥 cos𝑥] = 𝑥sin𝑥−1(sin 𝑥 + 𝑥 log 𝑥 cos 𝑥) …(ii)
and 𝑣 = (sin𝑥)cos𝑥 , taking log on both the sides
log 𝑣 = cos 𝑥 log sin 𝑥, therefore,
1
𝑣
𝑑𝑣
𝑑𝑥= cos 𝑥.
𝑑
𝑑𝑥log sin 𝑥 + log sin 𝑥.
𝑑
𝑑𝑥cos𝑥 = cos 𝑥.
1
sin𝑥cos 𝑥 + log sin𝑥(− sin 𝑥)
𝑑𝑣
𝑑𝑥= 𝑣[cos 𝑥 cot 𝑥 − sin 𝑥 log sin𝑥] = (sin 𝑥)cos𝑥(cos 𝑥 cot 𝑥 − sin𝑥 log sin 𝑥) …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (ii) in equation (i), we get
𝑑𝑦
𝑑𝑥= 𝑥sin𝑥−1(sin 𝑥 + 𝑥 log 𝑥 cos 𝑥) + (sin 𝑥)cos𝑥(cos 𝑥 cot 𝑥 − sin𝑥 log sin 𝑥)
-
Class-XII-Maths Continuity and Differentiabil ity
44 Practice more on Continuity and Differentiability www.embibe.com
10. Differentiate the functions given
𝑥𝑥 cos𝑥 +𝑥2+1
𝑥2−1
Solution:
Given function is 𝑥𝑥 cos𝑥 +𝑥2+1
𝑥2−1
Let 𝑢 = 𝑥𝑥 cos𝑥 and 𝑣 =𝑥2+1
𝑥2−1 therefore, 𝑦 = 𝑢 + 𝑣
Differentiating with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = 𝑥𝑥 cos𝑥 , taking log on both the sides
log 𝑢 = 𝑥 log 𝑥, therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥 cos 𝑥.
𝑑
𝑑𝑥log 𝑥 + log 𝑥.
𝑑
𝑑𝑥𝑥 cos 𝑥 = 𝑥 cos𝑥.
1
𝑥+ log 𝑥. (−𝑥. sin 𝑥 + cos 𝑥)
= cos 𝑥 − 𝑥 sin 𝑥 log 𝑥 + cos 𝑥 log 𝑥
𝑑𝑢
𝑑𝑥= 𝑢[cos 𝑥 − 𝑥 sin𝑥 log 𝑥 + cos 𝑥 log 𝑥]
= 𝑥𝑥 cos𝑥[cos 𝑥 − 𝑥 sin 𝑥 log 𝑥 + cos𝑥 log 𝑥] …(ii)
and 𝑣 =𝑥2+1
𝑥2−1, taking log on both the sides
log 𝑣 = log(𝑥2 + 1) − log(𝑥2 − 1), therefore,
1
𝑣
𝑑𝑣
𝑑𝑥=
1
𝑥2+1· 2𝑥 −
1
𝑥2−1· 2𝑥 =
2𝑥(𝑥2−1)−2𝑥(𝑥2+1)
(𝑥2+1)(𝑥2−1)=
−4𝑥
(𝑥2+1)(𝑥2−1)
𝑑𝑣
𝑑𝑥= 𝑣 [
−4𝑥
(𝑥2+1)(𝑥2−1)] =
𝑥2+1
𝑥2−1[
−4𝑥
(𝑥2+1)(𝑥2−1)] = −
4𝑥
(𝑥2−1)2 …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= 𝑥𝑥 cos𝑥[cos 𝑥 − 𝑥 sin 𝑥 log 𝑥 + cos𝑥 log 𝑥] −
4𝑥
(𝑥2−1)2
11. Differentiate the functions given
(𝑥 cos 𝑥)𝑥 + (𝑥 sin 𝑥)1
𝑥
Solution:
Given function is (𝑥 cos 𝑥)𝑥 + (𝑥 sin𝑥)1
𝑥
Let 𝑢 = (𝑥 cos 𝑥)𝑥 and 𝑣 = (𝑥 sin𝑥)1
𝑥, therefore, 𝑦 = 𝑢 + 𝑣
-
Class-XII-Maths Continuity and Differentiabil ity
45 Practice more on Continuity and Differentiability www.embibe.com
Differentiating with respect to 𝑥, we get
𝑑𝑦
𝑑𝑥=
𝑑𝑢
𝑑𝑥+𝑑𝑣
𝑑𝑥 …(i)
Here, 𝑢 = (𝑥 cos 𝑥)𝑥 , taking log on both the sides
log 𝑢 = 𝑥 log(𝑥 cos 𝑥), therefore,
1
𝑢
𝑑𝑢
𝑑𝑥= 𝑥.
𝑑
𝑑𝑥log(𝑥 cos𝑥) + log(𝑥 cos 𝑥).
𝑑
𝑑𝑥𝑥
= 𝑥.1
(𝑥 cos 𝑥)(−𝑥 sin 𝑥 + cos𝑥) + log(𝑥 cos 𝑥). 1 = −𝑥 tan 𝑥 + 1 + log(𝑥 cos 𝑥)
𝑑𝑢
𝑑𝑥= (𝑥 cos 𝑥)𝑥[1 − 𝑥 tan 𝑥 + log(𝑥 cos 𝑥)]
= (𝑥 cos 𝑥)𝑥[1 − 𝑥 tan 𝑥 + log(𝑥 cos 𝑥)] …(ii)
and, 𝑣 = (𝑥 sin 𝑥)1
𝑥, taking log on both the sides
log 𝑣 =1
𝑥log(𝑥 sin 𝑥) , therefore,
1
𝑣
𝑑𝑣
𝑑𝑥=1
𝑥.𝑑
𝑑𝑥log(𝑥 sin 𝑥) + log(𝑥 sin 𝑥).
𝑑
𝑑𝑥
1
𝑥
=1
𝑥.
1
𝑥 sin𝑥(𝑥 cos 𝑥 + sin𝑥) + log(𝑥 sin𝑥) (−
1
𝑥2)
𝑑𝑣
𝑑𝑥= 𝑣 [
𝑥 cot 𝑥 + 1 − log(𝑥 sin𝑥)
𝑥2]
= (𝑥 sin𝑥)1
𝑥 [𝑥 cot𝑥+1−log(𝑥 sin𝑥)
𝑥2] …(iii)
Putting the value of 𝑑𝑢
𝑑𝑥 from (ii) and
𝑑𝑣
𝑑𝑥 from (iii) in equation (i), we get
𝑑𝑦
𝑑𝑥= (𝑥 cos𝑥)𝑥[1 − 𝑥 tan 𝑥 + log(𝑥 cos 𝑥)] + (𝑥 sin 𝑥)
1
𝑥 [𝑥 cot𝑥+1−log(𝑥 sin𝑥)
𝑥2]
12. Find 𝑑𝑦
𝑑𝑥 of the functions given
𝑥𝑦 + 𝑦𝑥 = 1
Solution:
Given function is 𝑥𝑦 + 𝑦𝑥